# Define a linear transformation $T$, so that the null space is $z$-axis, and the range is the plane $x+y+z=0$

As stated in the title, it is requested to define a linear transformation $$T:\Bbb R^3 \to \Bbb R^3$$ such that the null space of $$T$$ is the $$z$$-axis, and the range of $$T$$ is the plane: $$x+y+z=0$$

I don't really know how to begin with the solution of the exercise, I think that I should try to get a matrix using the standard base, but after that, I don't have any concrete ideas.

You have to find a basis of that plane: $$x+y+z=0$$ then $$x=-y-z$$ so you can pick $$v_1=(1,-1,0), v_2=(1,0,-1)$$. The $$z$$-axis is the vector $$e_3=(0,0,1)$$.
$$f(e_1)=v_1, f(e_2)=v_2, f(e_3)=(0,0,0)$$
Choose a basis $$\mathcal{B}=\{u,v\}$$ for $$x+y+z=0$$ (for instance, it could be $$\{(2,-1,-1), (1,1,-2)\}$$).
Letting $$T$$ be the linear extension of $$\left\{\begin{array}{c} e_1\mapsto u \\ e_2 \mapsto v \\ e_3\mapsto 0\end{array}\right.$$ should work.